# How can I calculate the pH of a strong acid?

Mar 15, 2018

How else but by measurement? By definition, $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$

#### Explanation:

The $p H$ and $p O H$ scales were developed in the days before electronic calculators as a means to facilitate chemical calculations. And they used a logarithmic scale, the which chemists, and engineers, would have to look up, and manipulate.

In aqueous solution under standard conditions, we often speak of ${K}_{w}$, the ion product...where...

${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{- 14}$...

And we can take ${\log}_{10}$ of both sides to give....

${\log}_{10} {K}_{w} = {\log}_{10} {10}^{- 14} = {\log}_{10} \left[{H}_{3} {O}^{+}\right] + {\log}_{10} \left[H {O}^{-}\right]$.

And thus.... $- 14 = {\log}_{10} \left[{H}_{3} {O}^{+}\right] + {\log}_{10} \left[H {O}^{-}\right]$

Or.....

$14 = - {\log}_{10} \left[{H}_{3} {O}^{+}\right] - {\log}_{10} \left[H {O}^{-}\right]$

$14 = {\underbrace{- {\log}_{10} \left[{H}^{+}\right]}}_{p H} {\underbrace{- {\log}_{10} \left[O {H}^{-}\right]}}_{p O H}$

$14 = p H + p O H$

By definition, $- {\log}_{10} \left[{H}^{+}\right] = p H$, $- {\log}_{10} \left[H {O}^{-}\right] = p O H$